Problem: Solve for $x$ and $y$ using elimination. $\begin{align*}-8x+6y &= 5 \\ 7x-4y &= -6\end{align*}$
Explanation: We can eliminate $y$ when its corresponding coefficients are negative inverses. Recalling our knowledge of least common multiples, multiply the top equation by $2$ and the bottom equation by $3$ $\begin{align*}-16x+12y &= 10\\ 21x-12y &= -18\end{align*}$ Add the top and bottom equations. $5x = -8$ Divide both sides by $5$ and reduce as necessary. $x = -\dfrac{8}{5}$ Substitute $-\dfrac{8}{5}$ for $x$ in the top equation. $-8( -\dfrac{8}{5})+6y = 5$ $\dfrac{64}{5}+6y = 5$ $6y = -\dfrac{39}{5}$ $y = -\dfrac{13}{10}$ The solution is $\enspace x = -\dfrac{8}{5}, \enspace y = -\dfrac{13}{10}$.